Mathematics, 26.08.2019 02:10 xandraeden32

Since youtube first became available to the public in mid-2005, the rate at which video has been uploaded to the site can be approximated by v(t) = 525,600(0.42t2 + 2.7t − 1) hours of video per year (0.5 ≤ t ≤ 7), where t is time in years since the start of 2005.† use a definite integral to estimate the total number of hours of video uploaded from the start of 2006 to the start of 2011. (round your answer to the nearest million hours of video.)

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Answer from: Hellokittyjam35

$\displaystyle \int\limits^{6}_{1} {525600(0.42t^2 + 2.7t - 1)} \, dt \approx 38000000$

General Formulas and Concepts:

Calculus

Integration

Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle v(t) = 525600(0.42t^2 + 2.7t - 1) ,\ 0.5 \leq t \leq 7$

Step 2: Find

Set up: $\displaystyle \int\limits^{6}_{1} {525600(0.42t^2 + 2.7t - 1)} \, dt$ [Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^{6}_{1} {525600(0.42t^2 + 2.7t - 1)} \, dt = 525600\int\limits^{6}_{1} {(0.42t^2 + 2.7t - 1)} \, dt$ [Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int\limits^{6}_{1} {525600(0.42t^2 + 2.7t - 1)} \, dt = 525600 \bigg( \int\limits^{6}_{1} {0.42t^2} \, dt + \int\limits^{6}_{1} {2.7t} \, dt - \int\limits^{6}_{1} {1} \, dt \bigg)$ [Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^{6}_{1} {525600(0.42t^2 + 2.7t - 1)} \, dt = 525600 \bigg( 0.42\int\limits^{6}_{1} {t^2} \, dt + 2.7\int\limits^{6}_{1} {t} \, dt - \int\limits^{6}_{1} {} \, dt \bigg)$ [Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle \int\limits^{6}_{1} {525600(0.42t^2 + 2.7t - 1)} \, dt = 525600 \bigg[ 0.42 \bigg( \frac{t^3}{3} \bigg) \bigg| \limits^{6}_{1} + 2.7 \bigg( \frac{t^2}{2} \bigg) \bigg| \limits^{6}_{1} - t \bigg| \limits^{6}_{1} \bigg]$ Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^{6.5}_{1.5} {525600(0.42t^2 + 2.7t - 1)} \, dt = 525600 \bigg[ 0.42 \bigg( \frac{215}{3} \bigg) + 2.7 \bigg( \frac{35}{2} \bigg) - 5 \bigg]$ Simplify: $\displaystyle \int\limits^{6}_{1} {525600(0.42t^2 + 2.7t - 1)} \, dt = 38027160$ Round: $\displaystyle \int\limits^{6}_{1} {525600(0.42t^2 + 2.7t - 1)} \, dt \approx 38000000$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Answer from: Quest

D $e = mc\pi$

Answer from: Quest

You have 2 sides that are 2 cms and you have 6 sides that are 1/2 cm and you have 4 sides that are 1 cm. the smallest sides are 1/2 the medium sized sides are 1 and the smallest are 1/2. so to find the perimeter you have to add them all up together so your answer will be 11 cm's

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Since youtube first became available to the public in mid-2005, the rate at which video has been upl...